Solution: APPM 1360
Exam 2
Summer 2013
1. (20 pts) DO NOT SIMPLIFY YOUR ANSWERS FOR THIS PROBLEM.
p
Consider the region bounded by y = 1 − x2 , x = ey , y = −2 and the y-axis.
(a) Rotate this region about the line x = 1. Set up, but do not evaluate, an integral(s) for the volume
of the resulting solid using cylindrical shells.
(b) Rotate this region about the line x = 1. Set up, but do not evaluate, an integral(s) for the volume
of the resulting solid using disks/washers.
Solution: (a) Note√that x = ey → y = ln x. The region of interest goes from x = 0 (the y-axis) to the
intersection of y = 1 − x2 and y = ln x, which you cannot solve analytically; you must recognize that
the upper endpoint is x = 1. But the lower curve (for calculating the height of the cylindrical shells)
♥
changes from y = −2 to y = ln x where the two intersect: −2 = ln x → x = e−2 . Therefore...
p
p
For cylindrical shells: p
dV = 2πrh dr, where dr = dx, r = 1 − x, h = 1 − x2 − (−2) = 1 − x2 + 2 for
0 ≤ x ≤ e−2 , and h = 1 − x2 − ln x for e−2 ≤ x ≤ 1.
Z
Z e−2
Z 1
p
p
2
(1 − x)
→ V = dV = 2π
1 − x + 2 dx + 2π
(1 − x)
1 − x2 − ln x dx
e−2
0
(b) The region is between y = −2 (given) and y = 1 (where the circle intersects the y-axis).
For washers: dV = π(R2 − r2 )dh. dh = dy and note p
that the region must be broken up at y = 0 since
the inner radius changes from r = 1 − (ey ) to r = 1 − 1 − y 2 . The outer radius is always R = 1.
Z 1
Z
Z 0
2 p
2
2
y 2
2
(1) − 1 − 1 − y
dy
(1) − (1 − e ) dy + π
→ V = dV = π
0
−2
Z 0
Z 1
p
=π
2ey − e2y dy + π
y 2 − 1 + 2 1 − y 2 dy
−2
0
The region of interest is shown below.
π
.
3
(a) Set up but do not evaluate the integral(s) necessary to find the arc length of this curve between
these points.
2. (16 pts) Consider the curve f (x) = ln(cos x), 0 ≤ x ≤
(b) Set up but do not evaluate the integral(s) necessary to find the surface area of the solid resulting
from rotating this curve about the x-axis. Use vertical partitioning strips.
1
Solution: (a) f (x) = ln(cos x) → f 0 (x) =
(− sin x) = − tan x → 1 + (f 0 (x))2 = 1 + (− tan x)2 =
cos x
sec2 x, giving...
Z
L=
π
3
p
π
3
Z
1 + (f 0 (x))2 dx =
0
√
sec2 x dx =
Z
0
π
3
sec x dx
0
(b) dA = 2πrdL. Noting from the reminder given in the problem that f (x) < 0 for 0 ≤ x ≤
r = −f (x) = − ln(cos x). So:
Z π
3
A = 2π
(− ln(cos x)) sec x dx
π
3,
0
y
2
1
x
-9/4
-2
-1
1
-1/4
2
3. (14 pts) These nifty wine bottle holders use the concept of center-of-mass
to hover the bottle in mid-air. Approximate the holder as the region
between the lines y = 2, y = −x, y = 0 and y = −x − 41 , which is the
dashed red parallelogram in the figure. Approximate the wine bottle
as the rectangular region given by 1 ≤ y ≤ 2 and −1 ≤ x ≤ 2. Let ρ1
denote the uniform density of the holder and ρ2 denote the uniform
density of the wine bottle. As you might guess, the contraption will fall
over if the x-coordinate of its center of mass is not within the base of
the holder. Therefore, find an expression for ρ1 in terms of ρ2 which
will make the x-coordinate of the system’s center of mass be x̄ = − 18 .
m1 x̄1 + m2 x̄2
, where m1 and m2 are the masses of the holder and wine bottle,
m1 + m2
respectively, and x̄1 and x̄2 are the x-coordinates of the centers of mass of the holder and wine bottle.
Solution: The idea: x̄ =
From symmetry, we can see that x̄1 =
1
2
∗
−9
4
=
−9
8
and x̄2 =
1
2
∗ (2 − −1) = 12 .
Now the masses are m1 = ρ1 A1 and m2 = ρ2 A2 = 3ρ2 since the wine bottle is just a rectangle.
You may find the area of the holder in one of two ways:
(1) The area of a sheared rectangle (parallelogram) is the same as the original rectangle. Imagine grabbing
the top of the parallelogram and sliding over so the right side of it is flush with the y-axis. This creates
a normal rectange of area A1 = 41 ∗ 2 = 12 .
(2) The holder is the sum of 2 equal area triangles and the region between the lines y = −x and y = −x− 14
for −2 ≤ x ≤ −1
4 .
The triangles have base and height both equal to 14 since the lines are shifted by 41 from one another.
1
Thus the area of each triangle is 12 ( 14 )( 41 ) = 32
R −1
R −1
17
7
Between the lines, we have −24 ((−x) − (−x − 41 ))dx = −24 14 dx = 14 ( −1
4 − −2) = 4 4 = 16 .
1
+
Thus A1 = 2 32
7
16
=
8
16
= 12 .
♥
Now, plugging everything into the equation for x̄ = − 18 gives:
1 ♥ ρ1 ( 21 )(− 98 ) + ρ2 (3)( 12 )
1
3 ♥
9
3
8
♥ 15
♥ 15
− =
→ − ρ1 − ρ2 = − ρ1 + ρ2 → ρ1 = ρ2 → ρ1 = ρ2
8
16
8
16
2
16
8
4
ρ1 ( 12 ) + ρ2 (3)
dV
= V (−1 + cos t), where
4. (a) (8 pts) Suppose the volume of Kool-Aid at your party may be modeled as
dt
V is the volume of Kool-Aid at any given instant in time and t is time. If you and your roommates
initially bought 90 Kool-Aids, find an equation for your supply of Kool-Aid as a function of time.
dV
(b) (7 pts) Suppose, instead, you model your dwindling supply of Kool-Aid as
= 3t2 − 18t. Using the
dt
same initial condition as in (a), find a new equation for your supply of Kool-Aid as a function of time.
(c) (0 pts) Take a second to note that this is awesome! Now you can figure out whether your initial supply
of Kool-Aid will be sufficient, or if you’ll be making a Kool-Aid run later.
dV
dV
Solution: (a) separation of variables:
= V (−1 + cos t) →
= (−1 + cos t) dt →
dt
V
Z
(−1 + cos t) dt → ln |V | = −t + sin t + C → V (t) = e−t+sin t+C = e−t+sin t eC = Ae−t+sin t
Z
dV
=
V
Apply the initial condition, that V (0) = 90, to find V (0) = 90 = Ae0+sin 0 = A → A = 90. The answer is
then V (t) = 90e−t+sin t .
Z
Z
dV
2
2
(b) separation of variables again:
= 3t − 18t → dV = (3t − 18t)dt →
dV = (3t2 − 18t)dt →
dt
V (t) = t3 − 9t2 + C
apply the initial condition: V (0) = 90 = (0) − (0) + C → C = 90. the answer is then V (t) = t3 − 9t2 + 90.
(c) This sure is neat! Check out what the modeled Kool-Aid supply looks like for both of your potential
models:
5. (20 pts) Determine whether the following sequences converge or diverge. For those that converge, find
the limit. Justify your answers, show all work:
∞
n
o∞
1
n−1
n
(a)
(b)
(1
+
n)
n=1
(n2 + sin2 n) arctan(n) n=100
Solution: Put the Squeeze Theorem to work on this thing:
Since n starts at 100, we know that the numerator is certainly positive. Also: For 100 ≤ n < ∞,
arctan n ≥ arctan(100) > 0 and n2 + sin2 n > 0. Therefore:
n
n
n
n−1
≤
≤
≤ 2
=
0 ≤
2
2
2
2
2
2
n arctan(100)
(n + sin n) arctan(n)
(n + sin n) arctan(n)
(n + sin n) arctan(100)
1
n arctan(100)
1
and lim
=0
n→∞ n arctan(100)
n−1
therefore, lim
= 0 by the Squeeze Theorem.
n→∞ (n2 + sin2 n) arctan(n)
1
1
(b) an = (1 + n) n → ln(an ) = ln(1 + n), which is an indeterminant limit as n → ∞. Use L’Hopital’s
n
rule:
1
ln(1 + n) LH
= lim 1+n = 0
n→∞
n→∞ 1
n
→ lim ln(an ) = 0 → lim an = e0 = 1
lim
n→∞
n→∞
6. (15 pts) Answer either Always True or False. Do NOT justify your answer.
(a) If lim an = 2, then lim (a2n+1 − a2n ) = 0.
n→∞
n→∞
(b) If {an } and {bn } are divergent sequences, then the sequence {an + bn } is divergent.
dT
+ k(T − A) = 0, where k and A are positive
(c) T (t) = e−kt is a solution to the differential equation
dt
constants.
Solution: (a) TRUE: since an is convergent, all sub-sequences must also converge to the same limit, and
you can distribute the limit into the parentheses. So
lim (a2n+1 − a2n ) = lim a2n+1 − lim a2n = (2) − (2) = 0.
n→∞
n→∞
n→∞
(b) FALSE: Counterexample: an = (−1)n and bn = (−1)n+1 are both divergent, but adding them together
yields an + bn = 0 for all n, which is a convergent sequence.
(c) FALSE: take the derivative and plug it into the differential equation to see if the alleged solution works:
dT
dT
= −ke−kt →
+ k(T − A) = (−ke−kt ) + k((e−kt ) − A) = −kA 6= 0
dt
dt